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Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the number of noughts in
10 000! and 100 000! or even 1 000 000!

Find the largest integer which divides every member of the
following sequence: 1^5-1, 2^5-2, 3^5-3, ... n^5-n.

Number Rules - OK

Stage: 4 Challenge Level:

Solutions to parts, or all of this problem
have been received from Kirsty, Gilland Aimee, Natasha, Hamish,
Shona, Sheila and Alison Colvin all of Madras College, Maryof
Birchwood Community High School, Warrington, Andrei of School 205,
Bucharest, as well as Shu and Clement. Congratulations to all of
you for some very good solutions. Jack of Madras College started to
look at extensions to the pattern in part five of the question by
considering cubes of numbers. I think there might be a problem here
for a future edition of NRICH Jack!

Part 1

Can you convince me that if a square number is multiplied by a
square number the product is ALWAYS a square number?

Let the two square numbers be

$ a^2$ and $b^2$

$ a^2\times b^2 = a \times a \times b \times b =(ab)(ab) =
(ab)^2$

Therefore if a square number is multiplied by a square number the
product is ALWAYS a square number.

Part 2

Can you convince me that if a square number is divided by a
square number the product is ALWAYS a square number?

Therefore if a square number is divided by a square number the
quotient is ALWAYS a square number.

Part 3

Can you convince me that no number terminating in $2,3,7$, or
$8$ is a perfect square

A number ends in any of the digits $0-9$

When you square any number the final digit of the answer is
formed from the square of the units digit of the original
number.

Therefore the only possible endings for perfect squares are:

Ending in $1$:

$1^2$ or $9^2$

Ending in $4$:

$2^2$ or $8^2$

Ending in $5$:

$5^2$

Ending in $6$:

$4^2$ or $6^2$

Ending in $9$:

$3^2$ or $9^2$

So no number terminating in $2,3,7$, or $8$ is a perfect
square.

Part 4

Can you convince me that a number ending in $5$ cannot be a
perfect square unless the digit in the ten's column is a $2$.

Consider the square of a number ending in $5$ and with the tens
digits represented by the letter $b$.

The tens digit in the solution is formed from the units and tens
digits of number being mulitplied, plus anything that is carried
from the product of the units.

$a$

....

....

$b$

$5$

$\times$

$a$

....

....

$b$

$5$

$5a$

....

....

$5b^2$

$5$

$5b$

$0$

$0$

$0$

$0$

$0$

$0$

The units digit of the product =

$ 5 + 0 + 0+ \dots +0$

The tens digit of the product is the remainder when $2 + 5b + 5b +
0 + 0 + \dots + 0$ is divided by $10$ (the rest is caried into the
hundreds column). That is $2$ ($5b + 5b$ leaves a remainder of zero
when divided by $10$ )

Part 5

Convince me that the square of any number is equal to for
example:

$10^2 = 9^2 + 9 + 10$

Let any number be $n$. I will need to show that: $$n^2 = (n-1)^2
+ n-1 + n$$ $$n^2 = n(n-1)+n$$ $$\quad = (n-1)(n-1)+(n-1)+n$$
$$\quad = (n-1)^2+(n-1)+n$$ A useful representation of this result
was offered by Kirsty, Gill and Amy

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